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4.9t^2-5t-50=0
a = 4.9; b = -5; c = -50;
Δ = b2-4ac
Δ = -52-4·4.9·(-50)
Δ = 1005
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1005}=\sqrt{1*1005}=\sqrt{1}*\sqrt{1005}=1\sqrt{1005}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1\sqrt{1005}}{2*4.9}=\frac{5-1\sqrt{1005}}{9.8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1\sqrt{1005}}{2*4.9}=\frac{5+1\sqrt{1005}}{9.8} $
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